TITLE: Evaluate $\int_0^1 \frac{\operatorname{arctanh}^3(x)}{x}dx$
QUESTION [6 upvotes]: I'm trying to evaluate the following integral:
$$\int_0^1 \frac{\operatorname{arctanh}^3(x)}{x}dx$$
I was playing around trying to numerically approximate the answer with known constants and found that the integral is almost exactly $\frac{\pi^4}{64}$. The integral seems to break down after about $11$ decimal places.
I have a suspicion that this stems from the integral:
$$\int_0^1 \frac{\operatorname{arctanh}(x)}{x}dx$$
since this is exactly equal to $\frac{\pi^2}{8}$. 
Also for the integral:
$$\int_0^1 \frac{\operatorname{arctanh}^5(x)}{x}dx$$
This is suspicously close to $\frac{\pi^6}{128}$ but not exactly.
For some reason the above integrals diverges slightly from the some from of $\frac{\pi^n}{2^m}$ for some $n$ and $m$.

My question is: Why does this happen? And what are the true values of those integrals?

REPLY [3 votes]: An alternative route to the general result: For $z \in \mathbb{C}$ with $\operatorname{Re}(z) > 0$ we can let $x = \tanh(t/2)$ and use the geometric series to find
\begin{align}
\int \limits_0^1 \frac{\operatorname{artanh}^z (x)}{x} \, \mathrm{d} x &= \frac{1}{2^z} \int \limits_0^\infty \frac{t^z}{\sinh(t)} \, \mathrm{d} t = 2^{1-z} \sum \limits_{k=0}^\infty \int \limits_0^\infty t^z \mathrm{e}^{-(2k+1) t} \, \mathrm{d} t \\
&= 2^{1-z} \int \limits_0^\infty u^z \mathrm{e}^{-u} \, \mathrm{d} u \sum \limits_{k=0}^\infty \frac{1}{(2k+1)^{1+z}} = 2^{1-z} \Gamma(1+z) \lambda(1+z) \\
&= \frac{2^{1+z}-1}{4^z} \Gamma(1+z) \zeta(1+z) \, ,
\end{align}
where $\lambda$ is the Dirichlet lambda function.